Convolution Integrals — Question 7

Question 7 – Solution

Strategy. An unstable kernel remembers a weighted input area. Equal positive and negative ordinary areas need not cancel that memory.

Step 1: Factor the growing exponential. For an input supported in $[0,T]$, $$(h*f)(t)=e^t\int_0^{\min(t,T)}e^{-u}f(u)\,du.$$ Thus the output is zero for every $t\ge T$ if and only if $\boxed{\int_0^T e^{-u}f(u)\,du=0}$, since $e^t$ never vanishes.

Step 2: Distinguish the two cancellations. The ordinary area is $1-c$, so it vanishes at $c=1$. The weighted area is $(1-e^{-1})-c(e^{-1}-e^{-2})$; its unique zero is $\boxed{c=e}$. At $c=1$ it equals $(1-e^{-1})^2>0$, leaving a growing tail.

Step 3: Recover the selected response. Using the partial weighted integrals with $c=e$ gives $$\boxed{y_*(t)=\begin{cases}e^t-1,&0\le t<1,\\ e-e^{t-1},&1\le t<2,\\0,&t\ge 2.\end{cases}}$$ The pieces agree at both junctions. It increases up to $1$ and decreases from $1$ to $2$, so its unique maximum is $e-1$ at $t=1$.

Step 4: Track the transform domain. The convolution theorem initially gives, for $s>1$, $$Y(s)=\frac{1-(1+c)e^{-s}+ce^{-2s}}{s(s-1)}.$$ If $c\ne e$, the output has a nonzero $e^t$ tail and exact real domain $s>1$. If $c=e$, the output is compactly supported, so its transform exists for every real $s$; both apparent singularities at $0$ and $1$ are removable. The product formula agrees with it on $s>1$ and extends by removal of these singularities. Cancellation changes this output, not the unstable kernel. The dashed comparison uses $c=1$, which cancels only the ordinary input area.